Difference between revisions of "1953 AHSME Problems/Problem 50"
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Revision as of 11:32, 5 July 2013
Problem
One of the sides of a triangle is divided into segments of 
and 
units by the point of tangency of the inscribed circle. If the radius of the circle is 
, then the length of the shortest side is
Solution
Let the triangle have side lengths 
and 
. The area of this triangle can be computed two ways. We have 
, and 
, where 
is the semiperimeter. Therefore, 
. Solving gives 
as the only valid solution. This triangle has sides 
and 
, so the shortest side is 
.
See Also
| 1953 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 49 |
Followed by Last Question | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
